Is there a good errata for Atiyah-Macdonald available? A cursory Google search reveals a laughably short list here, with just a few typos. Is there any source available online which lists inaccuracies and gaps?

I think this question should be community wiki.
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Grétar AmazeenOct 15 '10 at 12:05

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I've just corrected the spelling of Ian G. Macdonald's name, to avoid confusion with the less famous group theorist Ian D. MacDonald. Aside from this, I think it's pointless to use this site to assemble errata for a book. The answer to the question about a Web source of errata is very likely no. If anybody wants to start a special Web site for this purpose, it's fine with me. Virtually all math books do have at least minor errors.
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Jim HumphreysOct 15 '10 at 23:19

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Let me try and "refute" the "high-profile organizer" comment above. To be frank, MO just turned out to be a repository for the Cassels-Froehlich errata rather than anything else. The reason I got so many was not because I posted here. It was because I asked for errata in many places rather than just here, all at the same time---but, crucially, I also approached several high-profile people personally (Hendrik Lenstra, Rene Schoof, J.-P. Serre, the Conrads [before, I think, they were MO-active] and others) and asked them if they had anything to send me...
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Kevin BuzzardOct 24 '10 at 9:25

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...and several responded with big lists. Note that almost all of the answers in that thread were posted by me, and are of the form "prof X just sent me this big list". I really pushed to make the errata, and, because I had a deadline myself (the LMS wanted to republish with the errata in) I had to push the people I was asking. I worked very hard to make those errata. So, it's very different to just posting once here and then sitting back and hoping (which, I think, is what is happening here, although I do apologise if I've got this wrong). Also C-F was typeset by a company who had very...
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Kevin BuzzardOct 24 '10 at 9:29

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...limited experience in mathematical typesetting, and they introduced many errors. (Oh---I should have mentioned Birch and Tate in my list of bigshots I approached directly, and I'm sure there are others I've forgotten).
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Kevin BuzzardOct 24 '10 at 9:29

22 Answers
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Dear Tim, on page 31 they consider a ring $A$ and two $A$- algebras defined by their structural ring morphisms $f:A\to B$ and $g:A\to C$. They then define the tensor product as a ring $D=B\otimes _A C$ and want to make it an $A$- algebra. For that they must define the structural morphism $A\to D$ and they claim that it is given by the formula $a \to f(a)\otimes g(a)$.This is false since that map is not a ring morphism. The correct structural map $A\to D$ is actually $a\mapsto 1_B\otimes g(a) =f(a)\otimes 1_C$.

PS: To prevent misunderstandings, let me add that Atiyah-MacDonald is, to my taste, the best mathematics book I have ever seen, all subjects considered.

Do you think we should create an errata here?
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Tim CampionOct 15 '10 at 16:13

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Dear Tim: no. I think we should stick to the community wiki format: one post per answer. This has the advantage that posts can then be commented (or refuted!) individually. Anyway, given my admiration for this almost perfect book, I conjecture that we will find very few errata ...
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Georges ElencwajgOct 15 '10 at 18:19

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Dear Tim, let me clarify my point of view. 1) I agree with you that there is probably no errata floating around. 2) I know no other error in the book and thus won't post any more. 3) I encourage everybody else to post new answers. 4) Afterwards, whoever wants to may create a document inspired by this community wiki and spread it as he/she deems fit.
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Georges ElencwajgOct 15 '10 at 21:00

@Yuji: Honorary titles in the British system are truly mysterious, but "Sir Atiyah" is actually "Sir Michael". Similarly, Swinnerton-Dyer is known as "Sir Peter". But lords and ladies are even more troublesome: the fictional detective is "Lord Peter" (not "Lord Wimsey") whereas the philosopher Bertrand Russell was "Lord Russell". Depends on birth order among other things, complicated by lifetime peerages for some. Feel free to call me Lord Jim....
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Jim HumphreysOct 17 '10 at 12:46

On page 29, the example at the top has two typos: it says "$(x)=2x$", when it should be "$f(x)=2x$", and the exact sequence at the end of that same line says "$0\rightarrow\mathbb{Z}\otimes \stackrel{f\otimes 1}{\longrightarrow} \mathbb{Z}\otimes N$", when it should be

"... $f$ induces a homomorphism $\widehat{f}:\widehat{G}\to\widehat{H}$, which is continuous."

No topology has been defined on $\widehat{G}$ and $\widehat{H}$.

[July 7, 2011, GMT. The topology on $\widehat{G}$ can be described as follows. For any subset $S$ of $G$, let $\widehat{S}\subset\widehat{G}$ be the set of equivalence classes of Cauchy sequences in $S$, and say that a subset $V$ of $\widehat{G}$ is a neighborhood of $0$ if there is a neighborhood $W$ of $0$ in $G$ such that $\widehat{W}\subset V$.]

By the way, there is (I think) a somewhat similar "mistake" in the article Atiyah wrote with Wall in "Algebraic Number Theory" Ed. Cassels and Froehlich (see Erratum for Cassels-Froehlich). Atiyah and Wall forgot to mention the crucial compatibility between change of groups and connecting morphisms. (See p. 99.)

Conversely, if $x\in r(\mathfrak a^e)$ then $x^n=\sum a_i\,x_i$ for some $n>0$, where the $a_i$ are elements of $\mathfrak a$ and the $x_i$ are elements of $C$. Since each $x_i$ is integral over $A$ it follows from (5.2) that $M=A[x_1,\dots,x_n]\ \dots$

It would be better (I think) to write something like

Conversely, if $x\in r(\mathfrak a^e)$ then $x^n=a_1\,x_1+\cdots+a_m\,x_m$ for some $m,n>0$, where the $a_i$ are elements of $\mathfrak a$ and the $x_i$ are elements of $C$. Since each $x_i$ is integral over $A$ it follows from (5.2) that $M=A[x_1,\dots,x_m]\ \dots$

[July 8, 2011, GMT. Page 90. It seems to me that the second part of the proof of Theorem 8.7 can be simplified. We must check the uniqueness of the decomposition of an Artin ring $A$ as a finite product of Artin local rings $A_i$. To do this it suffices to observe that, for each minimal primary ideal $\mathfrak q$ of $A$, there is a unique $i$ such that $\mathfrak q$ is the kernel of the canonical projection onto $A_i$.]

[July 9, 2015. The integer $d(M)$ (and in particular $d(A)$) is defined on p. 117 after the proof of Theorem 11.1. Another definition of $d(A)$ is given on p. 119 after the proof of Proposition 11.6 via the equality $d(A)=d(G_{\mathfrak m}(A))$. But the old meaning of $d(?)$ is used again in the proof of Proposition 11.20 p. 122, where the expression $d(G_{\mathfrak q}(A))$ occurs at the beginning of the last display. To avoid any confusion, let me denote by $D(M)$ the integer given by the first definition, and set $d(A):=D(G_{\mathfrak m}(A))$.

It seems to me the proof of Proposition 11.3 p. 118 is not entirely correct. I suggest to keep the proof, but to weaken slightly the statement, the new statement being: If $P(M/xM,t)\neq0$ and $D(M/xM)\ge1$, then $P(M,t)\neq0$ and $D(M/xM)=D(M)-1$.

This new statement applies to the first equality in the last display in the proof of Proposition 11.20 p. 122 if $d:=\dim A\ge1$ (the case $d=0$ being trivial). - On the third line of the proof $\mathfrak q$ should be $\mathfrak q^2$.]

And a semantic quibble rather than a mistake: I was confused for some time by the wording of exercise 5.2 on p.67 (if $A \subset B$ is an integral extension of rings, then any homomorphism of $A$ to an algebraically closed field extends to $B$). I've since concluded that A-M uses "homomorphism into" here to mean just "homomorphism to," i.e., not necessarily injective.
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Anna M. Oct 18 '10 at 17:52

On p.55, exercise 4.2 reads "If $\mathfrak a = r(\mathfrak a)$, then $\mathfrak a$ has no embedded prime ideals". I believe it should include the assumption that $\mathfrak a$ is decomposable.

A-M defines embedded primes for decomposable ideals only. And it doesn't seem that a radical ideal should automatically be decomposable. If you take something like a reduced (nonnoetherian) ring with infinitely many minimal prime ideals, I expect the zero ideal will be radical but not decomposable...

Dear Anna, you are exactly right. Take $A=k[x_1,x_2,...]=k[X_1,X_2,...]/I$ where $k$ is a field, $X_1,X_2,...$ are denumerably many indeterminates and $I=<X_i.X_j>_{i\neq j}$ is the ideal generated by the products of two different indeterminates. Then $<0>=\frak{a}=\sqrt \frak a$ is a reduced ideal which has no decomposition: indeed all the prime ideals $\sqrt {(0:x_i)}=<x_1,...,\hat x_i,...>=\frak p_i$ would have to be associated to a decomposition of $<0>=\frak a $ according to Theorem 4.5 and there are infinitely many such prime ideals: contradiction.
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Georges ElencwajgMay 1 '11 at 11:48

The way I saw that, either a) a is decomposable, and this makes sense or b) a is not decomposable, hence a fortiori has no embedded prime ideals!
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David CorwinAug 29 '12 at 7:59

Nearly all the mistakes pointed out so far were fixed in the Russian translation, which was done by Manin. But not all. I'll list in parentheses the page numbers of the translation where the original error still occurs for the 5 people who might care. (The translation is usually 11 page numbers ahead of the original.) Scan the answers posted before this one to determine which mistakes I am referring to.

p. 29 (---> p. 41): on line 8, change (2.14) to (2.13)

p. 55 (---> p. 66): exercise 2

p. 71 (---> p. 82): exercise 23

p. 88 (---> p. 99): exercise 27(v)

There were also completely original mistakes added especially for the translation!
On page 30 line -7 and page 31 lines 10 and 14 of the translation, the tensor product signs should be direct sum signs. On page 32 in the statement of Nakayama's Lemma, the ideal a should be in fraktur font.

Also minor: On p. 91, the $a$'s and $\mathfrak a$'s in the proof of Prop 8.8 seems to be a little jumbled.

I guess you want something like "Let $\mathfrak a$ be an ideal of $A$, other than $(0)$ or $(1)$. We have $\mathfrak m = \mathfrak N$, hence $\mathfrak m$ is nilpotent by (8.4) and therefore there exists a positive integer $r$ such that $\mathfrak a \subseteq \mathfrak m^r$ and $\mathfrak a \not\subseteq \mathfrak m^{r + 1}$; hence there exists $y \in \mathfrak a$ and $a \in A$ such that $y = ax^r$ but $y \not\in (x^{r + 1})$," etc.

"i) $\implies$ ii) by (3.5) and (2.20)" to "i) $\implies$ ii) by (3.7) and (2.19)"

On page 52 in remark 1) at the bottom of the page, change

"(see Chapter 1, Exercise 25)" to "(see Chapter 1, Exercise 27)"

On page 65 at the end of the proof of proposition 5.18. the black square to denote end of proof is missing.

On page 66 we need to correct the proof of corollary 5.22., one correct version is the following: We start with the quotient map $\pi: A[x^{-1}] \to A[x^{-1}] /m$ where $m$ is a maximal ideal containing $x^{-1}$. We take an algebraic closure $\Omega$ of the field $A[x^{-1}] /m$ and consider the map $i \circ \pi: A[x^{-1}] \to \Omega$. Then by the previous theorem, (5.21), we can extend $i \circ \pi$ to some valuation ring $B$ of $K$ containing $A[x^{-1}]$: $g: B \to \Omega$ such that $g|_{A[x^{-1}]} = i \circ \pi$. Then $g(x^{-1}) = 0$. Hence $x^{-1} \in ker(g)$ and since the kernel is a proper ideal of $B$, $x^{-1}$ is not a unit in $B$ and hence $x$ is not in $B$. (also see math.SE)

In p.45, Ex.3.12.iv, one can avoid the tedious argument provided in the hint by noting that $K\otimes_A M\cong(A-\{0\})^{-1}M$. (I originally did it as hinted ...)

In p.68, Ex.5.10.ii, (b') is actually equivalent to a weaker (c') that asserts only that $f^*:\mathrm{Spec}(B_\mathfrak{q})\to\mathrm{Spec}(A_\mathfrak{p})\cap V(\ker f)$ is surjective. However, (a') does imply the original (c').

page 20, line -12: in the expression for $A$ as a direct product, it should be $i=1$ not $i=I$.

page 28, line -5: $=$ should read $\cong$

page 29, line -13 (first line of proof that iv) $\Rightarrow$ iii)): $x_{i}$ should read $x_{i}'$

page 91, in the last example, it is not true that ${\mathfrak m}^{2}=0$. It is a non-zero principal ideal. But the following statement is still true, $\dim \left( {\mathfrak m}/{\mathfrak m}^{2} \right)=2$. It is generated by $x^2$ and $x^3$.

The part on page 69 is trivial, but the part on page 70 (still exercise 17) is the weak form of the Nullstellensatz.
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darij grinbergJun 14 '12 at 12:43

I think in Ex 16 and 17 one should replace $I(X)$ with an arbitrary ideal $I$ that defines $X$, and take $A=k[t_1,\cdots,t_n]/I$. The proof still goes through. This way the second part of Ex 17 can be derived from the first part by taking $I$ to be the maximal ideal.
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Junyan XuMar 5 '13 at 3:47

p. 82, definition of irreducible ideal: it is understood (by the first sentence of the section) that an irreducible ideal $\mathfrak{a}$ is also required to be proper (otherwise Lemma 7.12 would be false, for example). Also, there should perhaps be an entry "Ideal, irreducible, 82" in the index on p. 127.